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Why is this proposition always true?

$$\forall x\,\forall y\,\exists z\big((x<z)\to (x\ge y)\big)$$

And where`s the flaw in my thinking:

You can always choose a $z$ larger than $x$. So the problem can be reduced to:

For all $x$ and all $y$, it is such that $x\ge y$,

which obviously isn´t true...

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  • $\begingroup$ For instance, choose $z = x -1$. Then $(x < z)$ is always false and $(x < z) \to P$ is a true implication for any statement $P$. $\endgroup$
    – Simon S
    Oct 9, 2015 at 22:57
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    $\begingroup$ The proposition says that for any $x,y$ you can find a $z$ such that the latter statement is true. The latter statement is true if the '$P$' part is false. So you can pick $z=x$ and the statement $(x<z) \to (x \ge y)$ is true. $\endgroup$
    – copper.hat
    Oct 9, 2015 at 22:57

5 Answers 5

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The implication is equivalent to $(x\not<z)\lor(x\ge y)$. If $x\ge y$, you can pick any $z$, and this will be true. If $x<y$, just pick a $z$ such that $x\not<z$, and it will again be true. In fact, in all cases you can just pick $z=x$: either $x\ge y$, and the implication is true for that reason, or $x\not<x$, and the implication is true for that reason (or both, of course).

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And where`s the flaw in my thinking:

You can always choose a $z$ larger than $x$ So the problem can be reduced to:

For all x and all y , it is such that x≥y

There's the flaw. The implication being true does not mean its consequent is; especially if there is witness for $z$ which makes the antecedent false.

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The only way that $p\to q$ can be false is for $(\sim p)\wedge q$ to be true. So if $(x<z)\to (x\geq y)$ is false then $(x\geq z)\wedge (x<y)$ is true.Observe that $(x\geq z)\wedge (x<y)\iff (y>x\geq z).$ Now with $$p\iff (x<z)\text{ and }q\iff (x\geq y)$$ we have $$[\sim \forall x \forall y \exists z (p\to q)] \iff$$ $$[\exists x \exists y \forall z (\sim (p\to q))]\iff$$ $$[\exists x \exists y \forall z ((\sim p)\wedge q)]\iff$$ $$[\exists x \exists y \forall z (y>x\geq z)],$$ which implies the falsehood $$[\exists x \forall z (x\geq z)].$$ Your logical error was "you can always choose a $z$...".In the negation of the proposition, you cannot choose $z$ .It's "$\forall z$".

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Another way to see this: move the $z$ quantifier. The quantifier-free body of the statement is of the form $p \to q$ where $z$ is not free in $q$, so we can rewrite the statement as $\forall x \forall y ((\forall z p \to q)$:

$$ \forall x \forall y ((\forall z (x \lt z)\to(x \ge y)) $$ But $\forall z (x \lt z)$ is false, no matter what $x$ is, so for any $x, y$, $\forall z (x \lt z) \to(x \ge y)$ is "vacuously true" as it's equivalent to $0 \neq 0 \to(x \ge y)$. (something-false $\to$ anything) is always true!, given that $p \to q \iff \neg p \vee q$)

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Why is this proposition always true?

$\forall x\,\forall y\,\exists z\big((x<z)\implies (x\ge y)\big)$

Suppose we have $x$, $y$ in $\mathbb{R}$

By the irreflexive property of $\lt$, we have $\neg x\lt x$

Introducing $\lor$, we have $\neg x\lt x \lor x \ge y$

Applying the definition of $\implies $, we have $\neg\neg x\lt x \implies x\ge y$

Removing the double negation, we have $x\lt x \implies x\ge y$

Introducing $\exists$, we have $\exists z\in \mathbb{R}:[x\lt z \implies x\ge y]$

Generalizing, we have $\forall x,y \in \mathbb{R}: \exists z\in \mathbb{R}:[x\lt z \implies x\ge y]$

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